Here is a somewhat different way from Johan's of looking at this
problem. At each stage of the walk, choose a number $x$
uniformly from $[0,1]$ and then walk either a distance $x$ to the
right or $1-x$ to the left. This does not affect the probability
of becoming negative since there is still a uniform probability
of taking a step whose length belongs to the interval
$[-1,1]$. However, it does have the property that after taking
$n$ steps and choosing $0\leq x\leq 1$, the two possible
locations following the next step are the same modulo 1. Hence
the walk can be described as follows. Choose $n$ numbers
$0\lt x_1\lt \cdots\lt x_n\lt 1$, a sequence
$\epsilon=(\epsilon_1,\dots,\epsilon_n)$ of signs $\pm 1$, and a
permutation $w$ of $1,2,\dots,n$. Let the location be $y_k$ after
the $k$th step. If $\epsilon_k=1$ then step to the least real
number $y_{k+1}\equiv x_{w(k+1)}$ (mod 1), $y_{k+1}>y_k$. If
$\epsilon_k=-1$ then step to the greatest real number
$y_{k+1}\equiv x_{w(k+1)}$ (mod 1), $y_{k+1}\lt y_k$. But the
question of whether any $y_k$ is negative depends only on $\epsilon$ and
$w$, not the choice of $x_1,\dots,x_n$. There are $2^n n!$ ways
to choose $\epsilon$ and $w$. Is there a simple combinatorial
argument that the number of choices such that each $y_k>0$ is
$(2n-1)!!=1\cdot 3\cdot 5\cdots (2n-1)$? Then the probability of
success is $(2n-1)!!/2^nn! = (2n)!/4^nn!^2$.

Here is a reformulation of the combinatorial result that needs a
simple direct proof.

Let $f(n)$ be the number of pairs $(a_1a_2\cdots a_n,
b_1b_2\cdots b_{n-1})$ such that (a) $a_1 a_2\cdots a_n$ is a
permutation of $1,2,\dots, n$, (b) $b_i=0$ or $1$ if $a_i\lt
a_{i+1}$, (c) $b_i=0$ or $-1$ if $a_i>a_{i+1}$, and (d) $b_1
+b_2+\cdots+b_j\geq 0$ for all $1\leq j\leq n-1$. Then
$f(n)=(2n-1)!!$.

**Update.** The combinatorial result is proved bijectively by
O. Bernardi, B. Duplantier, and P. Nadeau in Séminaire
Lotharingien de Combinatoire, B63e (2010). In their citation [1]
they use this result for the same purpose as above, i.e., to
compute the probability $P_n$ (though they state the result a
little differently).

**Second update.** The method above can be applied to the $[l,r]$
generalization mentioned by Lwins in his comment. By rescaling we
may assume $l=-1$. If we are at $y$ sometime during the walk, choose
a number $x$ uniformly from $[0,1]$. With probability 1/2 step from
$y$ to $y+\frac{r-1}{2}+\frac{r+1}{2}x$. With probability 1/2 step
from $y$ to $y-1-\frac{r+1}{2}x$. This gives a uniform probability
of stepping from $y$ to a point in the interval $[y-1,y+r]$. It has
the property that once $x$ is chosen, the value of $y$ is determined
modulo $\frac{r+1}{2}$. Thus the walk can be described as follows:
pick uniformly and independently $0\lt x_1\lt \cdots\lt x_n \lt
\frac{r+1}{2}$,
pick a permutation $w$ uniformly from the symmetric group
$S_n$, and a sequence $\epsilon=(\epsilon_1,\dots,\epsilon_n)$ of
independently distributed signs, with a probability of
$\frac{r}{r+1}$ for a plus sign and $\frac{1}{r+1}$ for a minus sign.
Go through the same procedure as above, working mod
$\frac{r+1}{2}$ instead of mod 1. Again a proper walk (i.e., one
which never becomes negative) depends only on $w$ and $\epsilon$,
and we get the following result:

**Theorem.** The probability $P_n(r)$ that the walk is proper is
given by
$$ P_n(r) = \frac{1}{(1+r)^nn!}\sum r^{1+f(w,\beta)}, $$
summed over all pairs $w=a_1a_2\cdots a_n\in S_n$ and
$\beta=(b_1,\dots, b_{n-1})\in \lbrace 0,\pm 1\rbrace^n$ satisfying
the conditions (b) and (c) above, where $f(w,\beta)$ is the number
of integers $1\leq i\leq n-1$ for which either $a_i\lt a_j$ and
$b_i=0$, or $a_i\gt a_j$ and $b_i=1$.

For instance, $P_2(r)= (r+2r^2)/2(r+1)^2$ and $P_3(r)
=(r+8r^2+6r^3)/6(r+1)^3$. I conjecture that the numerator $N_n(r)$
of $P_n(r)$ is just the polynomial $\sum B_{n,i}r^i$ defined by
equation (4) of
http://math.mit.edu/~rstan/pubs/pubfiles/29.pdf. This paper gives
some additional information about the polynomials $\sum
B_{n,i}r^i$. Much additional information can be found in the
literature on Stirling permutations, e.g., Bona proves in
http://wenku.baidu.com/view/dfa70012cc7931b765ce15e4.html that all
zeros of this polynomial are real.

**Third update.** Alas, the conjecture in my second update is
false. Unless there is an error in my code, the sequence of
coefficients of $N_n(r)$ for $2\leq n\leq 7$ are $(1,2)$,
$(1,8,6)$, $(1,25,55,24)$, $(1,69,361,394,120)$,
$(1,176,1999,4416,3083,720)$,
$(1,426,9836,41019,52193,26620,5040)$. It is easy to see why
the leading coefficient of $N_n(r)$ is $n!$.